# how to find ionisation potential

how to find ionisation potential

## 2 Answers

**Actually ionisation potential means that "the amount of enrgy required to remove the electron from the valency shell of a particular atom".We have to find it as the depend upon its atomic size.It means that low size atoms have a low tendency to removing the electron.**

The Ionization Potentials and Shielding

of Electrons in the First Shell

The Bohr model of a hydrogen-like ion predicts that the total energy E of an electron is given by

E = −Z²R/n²

where Z is the net charge experienced by the electron, n is the principal quantum number and R is a constant

equal to approximately 13.6 electron volts (eV). This formula is the result of the total energy being

equal to

E = − Ze²/(2r

n

)

where e is the charge of the electron and r

n is the orbit radius when the principal quantum

number is n. The orbit radius is given by

r

n

= n²h/(Zm

e

e²)

where h is Planck''s constant divided by 2π and m

e is the mass of the electron.

Shell Structure

Electrons in atoms are organized in shells whose capacities are equal to 2m², where m is an integer. Thus there

can be at most 2 electrons in the first shell, 8 in the second shell and 8 in the third shell and 18 in each of the fourth

and fifth shells. Here only the first shell

is being considered. For elements above helium all of the electrons have been removed except two.

Here are all the ionization potentials for such ions.

The Ionization Potentials for the

Electrons

in the First Shell for the

Elements Helium through Copper

Proton

Number

Ionization

Potential

First

Electron

Ionization

Potential

Second

Electron

2 54.41778 24.58741

3 122.45429 75.64018

4 217.71865 153.89661

5 340.2258 259.37521

6 489.99334 392.087

7 667.046 552.0718

8 871.4101 739.29

9 1103.1176 953.9112

10 1362.1995 1195.8286

11 1648.702 1465.121

12 1962.665 1761.805

13 2304.141 2085.98

14 2673.182 2437.63

15 3069.842 2816.91

16 3494.1892 3223.78

17 3946.296 3658.521

18 4426.2296 4120.8857

19 4934.046 4610.8

20 5469.864 5128.8

21 6033.712 5674.8

22 6625.82 6249

23 7246.12 6851.3

24 7894.81 7481.7

25 8571.94 8140.6

26 9277.69 8828

27 10012.12 9544.1

28 10775.4 10288.8

29 11567.617 11062.38

The Ionization Potential of the First Electron

as a Function of Proton Number

An ion with only one electron is equivalent to the hydrogen atom but having a positive

charge of Z instead of one. The Bohr theory applies to such system. According to the

Bohr theory the ionization potential should be

IE = Z²R/n²

R is constant and n, the quantum number, is equal to 1. Thus the ionization potential should be proportional

to the proton number squared. Here is the plot of the relationship.

It certainly looks like a quadratic relationship. To test this proposition more precisely the logarithms of the

ionization potential and the proton numbers are computed. The plot of these quantitites is shown below.

The regression equation of the logartithm of ionization potential on the logarithm of the proton number is

ln(IE) = 2.604026184 + 2.004055992ln(#p)

[2029.1] [4123.2]

R² = 0.999998471

The numbers in the square brackets are the t-ratios for the regression coefficients. For a regression coefficient to

be statistically significantly different its magnitude must be greater than 2.0. As can be seen the regression coefficient for

ln(#p) is highly significant. And the value appears to be essentially 2, as the Bohr theory predicts. However this proposition

is tested by comparing the difference between the regression coefficient and 2 with the standard deviation of the regression

coefficient. Although the difference is small, 0.004056, the standard deviation of the regression coefficient is even smaller,

0.000486039, the ratio is 8.3. Common sense however says that the Bohr theory is verified for the first electron in the

the first shell.

The Z in the formula is the net charge experienced by the electron, the number of protons #p in the nucleus less any

shielding ε. The Bohr theory equation is then

IE = R(#p−ε)²

which can be expressed as

IE = R(#p² − 2#p*ε + ε²)

where R is the Rydberg constant, 13.6 eV.

Thus the appropriate regression equation would be

IE = c

0

+ c

1

#p + c

2

(#p)²

in which

c

1

<0

c

2

≅ R

Since in this case there seems to be no reason for there to be any shielding for the first electron the value of ε should be essentially zero.

The results of the regression are:

IE = 17.63611461 − 4.994593419#p + 13.89254281(#p)²

[5.1] [-9.7] [858.9]

R² = 0.999998221

The coefficient of #p is negative but statistically different from zero. The coefficient of (#p)² is notably close

to 13.6. The value of ε can be found as

ε = ½(−c

1

/c

2

) = 0.179758072

Somehow some small portion of the charge of the nucleus is being shielded.

The Bohr theory works marvelously well for an atom or ion with one electron but not nearly so well for multiple electrons.

One line of approach is to treat the atom or ion with two electrons as a many-body problem. This approach does not come up

with much in the way of definite results. An alternate line of approach is to say that electrons in the same shell shield

some fraction of a positive charge in the nucleus. This

shielding ratio would be based upon the proportion of the

charge of an electron which is closer to the nucleus than the center of an electron. The value of one half would be

plausible. However the shielding ratio could be affected by any deviation from spherical symmetry.

The Shielding Ratios of the Electrons in the First Shell

as a Function of Proton Number of the Nucleus

The shielding ratio ρ for the electrons in the first shell can be computed as

ρ = 1 − E

2

(#p)/E

1

(#p)

where E

1 and E

2 are the ionization potentials for the first and second electrons, respectively.

The values for the elements for which the data is available in the CRC Handbook of Physics and Chemistry

82nd Edition (2001-2002)

are given below.

The Shielding Ratio of

the Electrons

in the First Shell

for the Elements Helium through Copper

Proton

Number

Shielding

Ratio

2 0.548173226

3 0.382298652

4 0.293139977

5 0.237638033

6 0.199811573

7 0.172363225

8 0.151616443

9 0.135258834

10 0.122134019

11 0.111348807

12 0.10234044

13 0.09468214

14 0.088116709

15 0.082392514

16 0.077388254

17 0.072922812

18 0.068985102

19 0.065513374

20 0.062353287

21 0.059484443

22 0.056871451

23 0.054487091

24 0.052326782

25 0.050319998

26 0.048470039

27 0.046745345

28 0.045158416

29 0.043676844

The graph of these data displays a very regular pattern.

This appears to be a relationship of the form

ρ = α(#p)-β

The value of the parameters α and β can be found by plotting the logarithms of ρ and #p, as below.

A regression of ln(ρ) on ln(#p) gives:

ln(ρ) = 0.093012504 − 0.956463724ln(#p)

[16.2] [-439.2]

R² = 0.999865247

The results indicate that the shielding ratio is predictable on the basis of the proton number, but the shielding of positive charge in the nucleus by the electrons in the first shell

is primarily important for small nuclides.

The Ionization Potential for the Second

Electron as a Function of the Proton Number

Previously a regression equation of the form

IE = c

0

+ c

1

#p + c

2

(#p)²

was applied to the data for the first electron and it was found that there was a shielding of 0.179758072 units of charge.

When the data for the ionization potentials of the second electron is used the regression results are

IE = 20.41371337 − 21.54911763#p + 13.85998758(#p)²

[6.2] [-44.6] [912.9]

R² = 0.999998302

Again the coefficient of #p is negative and the coefficient of (#p)² is notably close to the Rydberg constant of 13.6.

The value of the shielding that is found from the coefficients is 0.777385892. The difference between this value and the

value for the first electron is 0.59762782, indicating that the second electron shields about 0.6 of a unit charge of the

nucleus. This is notably close to the value of 0.5 suggested by the simplest formulation of the notion of shielding by

electrons in the same shell.